Question

Refer the attached picture.

NUCLEI

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Answer 1

$\huge\underline\blue{\rm Answer:}$

$\large\red{\boxed{\sf Decay\: constant (\lambda)=0.23sec^{-1}}}$

$\large\red{\boxed{\sf Time (t)=12.45s}}$

$\huge\underline\blue{\rm Solution:}$

$\large\underline\pink{\sf Given: }$

• Half life ($\sf{t_{1/2}=30s})$

$\large\underline\pink{\sf To\:Find: }$

• Decay constant ($\sf{\lambda}$)=?

• Time taken for the sample to decay by 3/4th of the Initial value (t)= ?

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1. Decay constant ($\sf{\lambda}$)= ?

We know ,

$\Large{\boxed{\sf \lambda={\frac{0.693}{t_{1/2}}} }}$

$\large\implies{\sf \frac{0.693}{30}}$

$\large{\sf \lambda = 0.23 sec^{-1}}$

$\large\red{\boxed{\sf Decay\: constant (\lambda)=0.23sec^{-1}}}$

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2. Time taken for the sample to decay by 3/4th of the Initial value

We know ,

$\Large{\boxed{\sf N=N_o \left(\frac{1}{2}\right)^n}}$

$\large\implies{\sf \frac34N_o=N_o(\left\frac12\right)^n}$

$\large\implies{\sf \frac34=(\left\frac12\right)^n }$

Taking log on both side :

$\large{\sf log(\left\frac34\right)=log(\left\frac12\right)^n}$

$\large{\sf n=0.415 }$

Therefore ,

$\large{\boxed{\sf t=nt_{1/2} }}$

On Putting value

$\large{\sf t=0.415×30 }$

$\large{\sf t=12.45 }$

$\large\red{\boxed{\sf Time (t)=12.45s}}$

Time taken for the sample to decay by 3/4th of the initial value is 12.45s

Answer 2

Answer:-

(a) decay constant = 0.2 1/s

(b)T = 12.45 second .

Step - by - step explanation:-

Given that:-

$\: \bf{ t_{ \frac{1}{2} } = 30 \: seconds}$

Solution :-

(a) The decay constant =?

We know that,

$\: decay \: constant \: \lambda \: = \\ \implies \: \lambda \: = ? \frac{0.693}{ t_{ \frac{1}{2} } } \\ \\ \implies \: \lambda \: = \frac{0.693}{30} \\ \\ \implies \: \boxed{ \lambda \: = 0.23 \: {s}^{ - 1} }$

(b) time taken for the sample to decay by 3/4th of the initial value.

$\implies \: N \: = N_0 \: { \big( \frac{1}{2} \big) }^{n} \\ \\ \implies \: \frac{3}{4} N_0 = N_0 { \big(\frac{1}{2} \big) }^{n} \\ \\ \implies \: \frac{3}{4} = { \big(\frac{1}{2} \big) }^{n} \: \\ \\ taking \: log \: on \: both \: sides \\ \implies \: log( \frac{3}{4} ) = n log( \frac{1}{2} ) \\ \\ \implies \: \boxed{n = 0.415}$

Therefore,

T = 0.415 × 30

T = 12.45 second .